Stock market puzzles :) - page 3

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TEXX >>:

Ну в книжке основных 13 написано...

Ну и сколько у Вас?

Well, is there any proof (any proof at all) that there are that many?

 
SProgrammer писал(а) >>

Well, what's the proof (any proof at all) that that's how many there are?

I don't feel like drawing:)

But I'm sticking with 10.

 
TEXX >>:

Рисовать неохота:)

Но я придерживаюсь 10.

You have to prove it, not draw it. Or to avoid going into the maze of topology, you could just write a program that classifies the bar.

 

if you mean by HLOC attitude, 12 - by enumeration

H=O - 2pc

H=C - 2pc

L=O - 2pc

L=C - 2pc

HOCL and HCOL and O=C 3 more

and when all at the same level H=O=C=L - 1pc

-------------------

P.S. I recalculated, it turned out to be 10 as some combinations (H=O, L=C and H=C, L=O) counted twice :)

 
Avals >>:

если вы имеете в виду по отношение HLOC, то 12 - перечислением

H=O - 2шт

H=С - 2шт

L=O - 2шт

L=C - 2шт

HOCL и HCOL и O=C еще 3

и когда все на одном уровне H=O=C=L - 1шт

-------------------

P.S. пересчитал, оказалось 10 т.к. некоторые комбинации (H=O, L=C и H=C, L=0) учел дважды :)

Are you sure (I'm just asking you to give a proof, because how else would you do it) that you have accounted for all the options?

The easiest way to "prove" is to write 15 lines in MQL and clearly and without misunderstanding.

Let's say there is a function 

int BType ( int n )
{
   double H=High[ n], L=Low[ n], C=Close[ n], O=Open[ n];

  // тут вычисляем тип как int
  
  return t;
}

***

 
SProgrammer писал(а) >>

Are you sure (I'm just asking you to give a proof, because how else would you do it) that you have considered all the options?

The easiest way to "prove" is to write 15 lines in MQL and clearly and without misunderstanding.

Let's say there is a function

The program is not a proof - you'll get what you need in the logic. In kotirs there may be a glitch when an open or a close is above the high or below the low, it will not prove that these variants are correct. Although it should be a good option for MT.

 
Avals >>:

программа не является доказательством - что задаш в логике, то и получешь. Ну найдется глюк в котирах, когда опен или клозе находятся выше хая или ниже лоу, это ведь не докажет что это правильные варианты. Хотя такое МТ вроде должен отсекать

For example


if ( a > 0 )

sss;

else

fff;


There is a proof that there are only two options here.

The program is an algorithm - when you look at it, it's immediately clear if the solution is right or wrong. :)

 
SProgrammer писал(а) >>

For example

if ( a > 0 )

sss;

else

fff;

There is a proof that there are only two options here.

The program is an algorithm - when you look at it, it's immediately clear if the solution is right or wrong. :)

Too lazy to write, especially on a Saturday.

But still I "vote" for 12, because I didn't consider variants H=C=O and L=C=O.

 
Avals >>:

лень писать, тем более в субботу

Но все же я "голосую" за 12, т.к. не учел варианты H=C=O и L=C=O

You see, it's not as easy as it looked :) in the beginning.

 
And it's precisely as an algorithmic one that makes it even more interesting. :)
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